*Originally posted by David113*

**This is not a proof. How do you know that if the strips are not all parallel, and the sum of their widths is less than D, then their union (the area covered by at least one strip) can't cover a disc of diameter D?
**

This problem DOES have a nice proof...

The best way to lay the strips in an abstract world:

Assume two strips are not parallel, and thus they overlap.

Notice that the maximum area that any two strips of width a and b cover is when they are laid parallel and side by side with a diameter through (a+b)/2 (the centre of the "new" strip of width a+b.

Then the area of the circle that they cover is STRICTLY less than the maximum area of the circle that they cover (when they are not parallel).

Thus, parallel is obviously the best way to lay the first two strips. So, create a strange almost ellipse like shape, but with pointed corners, by removing the part of the circle covered by placing the two strips of largest width in the best starting way then removing these strips and the part of the circle that they cover and amalgamating the two parts of the circle left.

Now, applying the same logic as before we can see that the best way too lay any two strips is through the centre as before and along the "wider" part of this ellipse-esq shape.

Apply until we have used all the strips.

What we have done has not changed the total width of the strips, we have just sort of displaced the way in which the width has been distributed. (Two strips of width a and b now each have with (a+b)/2 if they were two strips that we laid in the same "manouver", but (a+b)/2+(a+b)/2=a+b, and thus the total width of the strips we have lain is equal to the total width of the proper strips).

Now, the way we have lain these strips covers more area on the circle than any other method of laying the strips. However, it does not cover the circle as the total width is less than D.

Thus, it is impossible.